For any linear transformation we define the norm of it as [itex]||T||=\sup_{|x|<1}|T(x)|[/itex]. Your arguments show that this norm always exists for any linear map.

Then we have [itex]|T(x-y)|\le||T|||x-y|[/itex]. Therefore [itex]T[/itex] is Lipschitz, hence uniformly continuous. If you don't know anything about Lipschitz and uniform continuity you can just let [itex]|x-y|<\epsilon/||T||[/itex] and we have our desired result.

1.Thanks for pointing that out, it should be the absolute value.
2.sorry i made a mistake there, it should be
sum_i|xi|<sqrt(m)*sqrt(sum_i(xi^2)) by AM<RMS

Oh, ok. I did not know this inequality, but it is true as long as you change that < for a <=, because obviously for m=1, we have an equality. What does AM and RMS stand for?

What you've proven in showing |T(x)|<=C|x| is that T is Lipschitz continuous at 0. Usually, we say that a function f is Lipschitz at x0 if there exists numbers c and M such that |x-x0|<c ==> |f(x)-f(x0)| [itex]\leq[/itex] M|x-x0|. A function that is Lipschitz is necessarily continuous because given e>0, choose d=min{c,e/M}.

So, can you extend your result and show that T is Lipschitz everywhere?